Question 109 of 180

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants ${K}_{1},{K}_{2},{K}_{3},{K}_{4}$ arranged as shown in the figure. The effective dielectric constant $K$ will be:

A$K=\left(\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}+\dfrac{{K}_{3}.{K}_{4}}{{K}_{3}+{K}_{4}}\right)$B$K=\dfrac{\left({K}_{1}+{K}_{2}\right)\left({K}_{3}+{K}_{4}\right)}{2\left({K}_{1}+{K}_{2}+{K}_{3}+{K}_{4}\right)}$C$K=\dfrac{({K}_{1}+{K}_{4})({K}_{2}+{K}_{3})}{2{\left({K}_{1}+{K}_{2}+{K}_{3}+{K}_{4}\right)}_{}}$D$K=\dfrac{\left({K}_{1}+{K}_{2}\right)({K}_{3}+{K}_{4})}{{K}_{1}+{K}_{2}+{K}_{3}+{K}_{4}}$

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Answer: (A) $K=\left(\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}+\dfrac{{K}_{3}.{K}_{4}}{{K}_{3}+{K}_{4}}\right)$

${C}_{1}$ and ${C}_{2}$ are in series.

The equivalent capacitance of ${C}_{1}$ and ${C}_{2}$ is

${C}_{12}=\dfrac{{\in }_{0}\dfrac{A}{2}}{\dfrac{d}{2{K}_{1}}+\dfrac{d}{2{K}_{2}}}=\dfrac{{\in }_{0}A}{d}\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}$
${C}_{3}\text{and}{C}_{4}\text{are in series}$

${C}_{34}=\dfrac{{\in }_{0}\dfrac{A}{2}}{\dfrac{d}{2{K}_{3}}+\dfrac{d}{2{K}_{4}}}=\dfrac{{\in }_{0}A}{d}\dfrac{{K}_{3}{K}_{4}}{{K}_{3}+{K}_{4}}$

${C}_{eq}={C}_{12}+{C}_{34}=\dfrac{{\in }_{0}A}{d}\left(\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}+\dfrac{{K}_{3}{K}_{4}}{{K}_{3}+{K}_{4}}\right)$

$∴{K}_{eq}=\left(\dfrac{{K}_{1}{K}_{2}}{{K}_{1}+{K}_{2}}+\dfrac{{K}_{3}.{K}_{4}}{{K}_{3}+{K}_{4}}\right)$